$f$ is the Maclaurin series $\sum_{n=0}^{\infty}\frac{{{x}^{n}}}{2n!}$. $f\left(\ln(7)\right)=$
Note that the terms have $~n!~$ in the denominator and that they do not alternate in sign. That suggests that we use the Maclaurin series for $~{{e}^{x}}~$ as a guide. Recall that ${{e}^{x}}=\sum_{n=0}^{\infty}\frac{{{x}^{n}}}{n!}\,$. We see that our series is $\sum_{n=0}^{\infty}\frac{{{x}^{n}}}{2n!}\,=\,\frac{1}{2}\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=\,\frac{1}{2}e^x$. Now substitute $x = \ln(7)$. $\frac{1}{2}e^x = \frac{1}{2}e^{\ln(7)}=\frac{7}{2}\,$.